Retract of a quasi-isomorphism of chain complexes.
$require{AMScd}$
I want to show that the retract of a quasi-isomorphism of chain complexes is also a quasi-isomorphism of chain complexes. Let $X,Y,A,B$ be chain complexes in degrees greater than and equal to $0$. Say that $f:Xto Y$ is a qis, and $g:Ato B$ is a retract of $f$:
$$begin{CD}
A@>>> X@>>>A\
@VVV @VVV @VVV\
B @>>> Y@>>> B
end{CD}$$
where the top composite is the identity, the bottom composite is the identity, $Xto Y$ is a qis and the diagram commutes.
Then taking homology and considering any degree $n$ we have:
$$begin{CD}
H_n(A)@>>> H_n(X)@>>>H_n(A)\
@VVV @VVV @VVV\
H_n(B) @>>> H_n(Y)@>>> H_n(B)
end{CD}$$
- The top row is the identity composition,
- Thus $H_n(A)to H_n(X)$ is injective,
- Then the composite $H_n(A)to H_n(B)to H_n(Y)$ is injective,
- Thus the left square has $H_n(A)to H_n(B)$ injective.
- Similarly $H_n(Y)to H_n(B)$ is surjective.
- Since the right square commutes we have that $H_n(A)to H_n(B)$ is surjective.
- But the injective map on the left column is the same map $g_*$ as the surjective map on the rightmost column.
- Then $H_n(A)to H_n(B)$ is an isomorphism.
- This occurs in each degree and hence $Ato B$ is a qis.
Is this correct?
abstract-algebra homology-cohomology homological-algebra
asked Dec 8 at 16:47
user616128
405
add a comment |
$require{AMScd}$
I want to show that the retract of a quasi-isomorphism of chain complexes is also a quasi-isomorphism of chain complexes. Let $X,Y,A,B$ be chain complexes in degrees greater than and equal to $0$. Say that $f:Xto Y$ is a qis, and $g:Ato B$ is a retract of $f$:
$$begin{CD}
A@>>> X@>>>A\
@VVV @VVV @VVV\
B @>>> Y@>>> B
end{CD}$$
where the top composite is the identity, the bottom composite is the identity, $Xto Y$ is a qis and the diagram commutes.
Then taking homology and considering any degree $n$ we have:
$$begin{CD}
H_n(A)@>>> H_n(X)@>>>H_n(A)\
@VVV @VVV @VVV\
H_n(B) @>>> H_n(Y)@>>> H_n(B)
end{CD}$$
- The top row is the identity composition,
- Thus $H_n(A)to H_n(X)$ is injective,
- Then the composite $H_n(A)to H_n(B)to H_n(Y)$ is injective,
- Thus the left square has $H_n(A)to H_n(B)$ injective.
- Similarly $H_n(Y)to H_n(B)$ is surjective.
- Since the right square commutes we have that $H_n(A)to H_n(B)$ is surjective.
- But the injective map on the left column is the same map $g_*$ as the surjective map on the rightmost column.
- Then $H_n(A)to H_n(B)$ is an isomorphism.
- This occurs in each degree and hence $Ato B$ is a qis.
Is this correct?
abstract-algebra homology-cohomology homological-algebra
asked Dec 8 at 16:47
user616128
405
add a comment |
$require{AMScd}$
I want to show that the retract of a quasi-isomorphism of chain complexes is also a quasi-isomorphism of chain complexes. Let $X,Y,A,B$ be chain complexes in degrees greater than and equal to $0$. Say that $f:Xto Y$ is a qis, and $g:Ato B$ is a retract of $f$:
$$begin{CD}
A@>>> X@>>>A\
@VVV @VVV @VVV\
B @>>> Y@>>> B
end{CD}$$
where the top composite is the identity, the bottom composite is the identity, $Xto Y$ is a qis and the diagram commutes.
Then taking homology and considering any degree $n$ we have:
$$begin{CD}
H_n(A)@>>> H_n(X)@>>>H_n(A)\
@VVV @VVV @VVV\
H_n(B) @>>> H_n(Y)@>>> H_n(B)
end{CD}$$
- The top row is the identity composition,
- Thus $H_n(A)to H_n(X)$ is injective,
- Then the composite $H_n(A)to H_n(B)to H_n(Y)$ is injective,
- Thus the left square has $H_n(A)to H_n(B)$ injective.
- Similarly $H_n(Y)to H_n(B)$ is surjective.
- Since the right square commutes we have that $H_n(A)to H_n(B)$ is surjective.
- But the injective map on the left column is the same map $g_*$ as the surjective map on the rightmost column.
- Then $H_n(A)to H_n(B)$ is an isomorphism.
- This occurs in each degree and hence $Ato B$ is a qis.
Is this correct?
abstract-algebra homology-cohomology homological-algebra
asked Dec 8 at 16:47
user616128
405
$require{AMScd}$
I want to show that the retract of a quasi-isomorphism of chain complexes is also a quasi-isomorphism of chain complexes. Let $X,Y,A,B$ be chain complexes in degrees greater than and equal to $0$. Say that $f:Xto Y$ is a qis, and $g:Ato B$ is a retract of $f$:
$$begin{CD}
A@>>> X@>>>A\
@VVV @VVV @VVV\
B @>>> Y@>>> B
end{CD}$$
where the top composite is the identity, the bottom composite is the identity, $Xto Y$ is a qis and the diagram commutes.
Then taking homology and considering any degree $n$ we have:
$$begin{CD}
H_n(A)@>>> H_n(X)@>>>H_n(A)\
@VVV @VVV @VVV\
H_n(B) @>>> H_n(Y)@>>> H_n(B)
end{CD}$$
- The top row is the identity composition,
- Thus $H_n(A)to H_n(X)$ is injective,
- Then the composite $H_n(A)to H_n(B)to H_n(Y)$ is injective,
- Thus the left square has $H_n(A)to H_n(B)$ injective.
- Similarly $H_n(Y)to H_n(B)$ is surjective.
- Since the right square commutes we have that $H_n(A)to H_n(B)$ is surjective.
- But the injective map on the left column is the same map $g_*$ as the surjective map on the rightmost column.
- Then $H_n(A)to H_n(B)$ is an isomorphism.
- This occurs in each degree and hence $Ato B$ is a qis.
Is this correct?
abstract-algebra homology-cohomology homological-algebra
abstract-algebra homology-cohomology homological-algebra
asked Dec 8 at 16:47
user616128
405
asked Dec 8 at 16:47
user616128
405
asked Dec 8 at 16:47
user616128
405
asked Dec 8 at 16:47
user616128
405
asked Dec 8 at 16:47
user616128
405
405
add a comment |
add a comment |
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3031339%2fretract-of-a-quasi-isomorphism-of-chain-complexes%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3031339%2fretract-of-a-quasi-isomorphism-of-chain-complexes%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
